MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be a commutative noetherian ring. I know that an $R$-module is invertible iff it is finitely generated and locally free of rank one. I presume then that there are examples of non-finitely generated, rank one projective modules which are not invertible. Can someone help me out by providing a specific example? Additionally, is there any extension of the notion of the Picard group that includes non-finitely generated rank one projectives?

share|cite|improve this question
How do you define the rank in this case? Dimension of the residue fields? Of course your alluded generalized Picard group will be no group. – Martin Brandenburg Jul 2 '11 at 19:40
Dear Karl, I am not able to extract an answer to Andrew's question from your link. Could you please state a precise statement there that answers his question? – Georges Elencwajg Jul 3 '11 at 8:23
Andrew, what definition of "invertible" you are using? The ones I know would force $M$ to be finitely generated automatically, so the answer is trivially no. – Hailong Dao Jul 3 '11 at 15:51
@Hailong: the actual question does not ask $M$ to be invertible, in fact it asks that it is not that, so it seems that that could not force anything. But it does ask that $M$ be of rank $1$ and perhaps this is what you have in mind. So my question is the same as Martin's: @Andrew: How do you define rank? – Sándor Kovács Jul 3 '11 at 17:32
@S\andor: you are right, I meant to say, if $M$ is not f.g. (as the OP wanted), then it is trivially not invertible. So I am not sure what the question asked. Thank you for the correction. – Hailong Dao Jul 3 '11 at 18:41
up vote 10 down vote accepted

A rank one projective module $M$ over a commutative noetherian ring is necessarily finitely generated. Indeed assume otherwise. Let $a_i$ be the minimal idempotents of $R$ (there are finitely many since $R$ is commutative noetherian), so that $R=\bigoplus_i a_iR$. Then $a_iM$ is projective over the connected (=indecomposable) noetherian ring $a_iR$. Bass (Illinois Math J, 1963) showed that a infinitely generated projective module over a connected noetherian commutative ring is necessarily free. So here $a_iM$ is in addition free of rank one over $a_iR$. If follows that $M=\bigoplus_i a_iM$ is finitely generated (actually: free of rank one), contradiction.

Remarks: 1) I assume any reasonable definition of "[locally free of] rank one", for instance the (weak) assumption that $M\otimes R_P$ is free of rank one for every prime $P$ in $R$.

2) Tom Goodville gave a nice example of a module, locally free of rank one, that is not projective. [Recall that a f.g. locally free module has to be projective.] This is the $\mathbf{Z}$-submodule of $\mathbf{Q}$ consisting of fractions $a/b$ with squarefree $b$. It is locally free in the above weak sense (localization at primes), but not in the stronger sense, where the strong sense of locally free would be: free over every open subset in some open covering of the spectrum (is this equivalent to being projective? I'm not sure in the infinitely generated case).

share|cite|improve this answer
@Yves, thanks for your clarification re projectivity & FG. I was considering exactly what you are referencing in your first remark, that $M \otimes R_P \equiv R_P$ for all primes $P$ as my 'definition' for invertibility. – Andrew Parker Jul 5 '11 at 17:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.